PeroK said:It is important, though, to realize that a limit is a number, and not a process. It's a small point, perhaps, but it sharpens your understanding of these things.
madness said:Yes, but it's also important to remember that this number (the limit) is the number that you get closer and closer to as you apply the process of increasing n. Or, if you don't like this process, it's the number that you can get as close to as you like by choosing a high enough value of n.
Neither is a rigorous definition of a limit but they are a statement of what is implied by such a definition. In particular, the limit of the sequence 0.9, 0.99, 0.999, ... is the number 1, not the sequence itself.PeroK said:Neither of those is the definition of a limit.
HallsofIvy said:Neither is a rigorous definition of a limit but they are a statement of what is implied by such a definition. In particular, the limit of the sequence 0.9, 0.99, 0.999, ... is the number 1, not the sequence itself.
The phrasing that madness used was careful and avoided speaking of the result of a never-ending process.PeroK said:Perhaps, but my original point was that there are those who see the limit as a number (full stop) and those who see the limit as the hypothetical result of a never-ending process. Breaking away from this latter view is perhaps an important step in grasping rigorous maths.
madness said:Yes, but it's also important to remember that this number (the limit) is the number that you get closer and closer to as you apply the process of increasing n. Or, if you don't like this process, it's the number that you can get as close to as you like by choosing a high enough value of n.
jbriggs444 said:The phrasing that madness used was careful and avoided speaking of the result of a never-ending process.
Mark44 said:We are not talking about a specific Riemann sum. My complaint is that you are talking about a limit, but don't seem to realize it. Emphasis added.
In other words, as ##\Delta x \to 0## -- that's a limit. All the other stuff about compact sets, continuity, and subcovers etc. is just word salad that I'm 99% sure would be over the head of the person you were responding to.
I think you mean Riemann integral being the limit of the Riemann sums.WWGD said:No, I think you are not realizing my point. I am talking about an individua Riemann sum not being 0, addressing the question. I do know that the Riemann sum is the limit ( if it exists) of the Riemann sums.
WWGD said:I never said nor implied otherwise. I did reply to the OP that for individual sums, the width is controlled, so it does not go to zero. The rest of the exchanges were with and for you.
The rectangle widths are either going to zero or they're not -- you can't have it both ways.WWGD said:I think the point is that the rectangles, while going to 0, are uniformly bounded , by compactness of the region of integration, so these widths do not become indefinitely-close to 0
Mark44 said:I think you mean Riemann integral being the limit of the Riemann sums.My complaint is in the following. I interpret the phrase "while going to 0" as equivalent to saying "the limit of the Riemann sums as n goes to infinity." You can't say that the widths of the rectangles are going to zero, but at the same time do not become close to zero. Here again is what you said, with emphasis added.
The rectangle widths are either going to zero or they're not -- you can't have it both ways.
When you say, "the rectangles, while going to 0, ..." you're no longer talking about a specific Riemann sum -- you're talking about the limit of the Riemann sums (as the number of terms goes to infinity), which means you're talking about the Riemann integral.WWGD said:I was referring to individual Riemann sums, not to the Riemann integral itself. I was addressing the question of whether a Riemann sum is zero or not, because the general term goes to zero in an uncontrolled way.
In short, you're saying that the rectangles are going to zero and that their widths are not going to zero. That's impossible.I think the point is that the rectangles, while going to 0, are uniformly bounded , by compactness of the region of integration, so these widths do not become indefinitely-close to 0
WWGD said:I made no reference to the Riemann integral itself; that is a whole other topic. But yes, the Riemann integral is the limit as the partition width goes to zero. I don't see why the two contradict or are incompatible with each other.
Mark44 said:When you say, "the rectangles, while going to 0, ..." you're no longer talking about a specific Riemann sum -- you're talking about the limit of the Riemann sums (as the number of terms goes to infinity), which means you're talking about the Riemann integral.
These were your exact words:
In short, you're saying that the rectangles are going to zero and that their widths are not going to zero. That's impossible.
Also, I don't know what you mean "goes to zero in an uncontrolled way" ...
Please explain to me how the widths of the rectangles can be "going to 0" and at the same time "these widths do not become indefinitely-close to 0" -- your words.WWGD said:No, I am definitely talking about an individual Riemann sum. By a controlled way, I mean there is a lower bound for the widths, which there is by the usual arguments. I think the confusion here is you are referring to the Riemann integral, while I am referring to specific Riemann sums.
Mark44 said:We are not talking about a specific Riemann sum. My complaint is that you are talking about a limit, but don't seem to realize it. Emphasis added.
In other words, as ##\Delta x \to 0## -- that's a limit. All the other stuff about compact sets, continuity, and subcovers etc. is just word salad that I'm 99% sure would be over the head of the person you were responding to.
Mark44 said:Please explain to me how the widths of the rectangles can be "going to 0" and at the same time "these widths do not become indefinitely-close to 0" -- your words.
I maintain that you can't possibly be talking about a specific Riemann sum with a specific number of terms (hence a specific number of rectangles, all of whose areas are positive numbers).
Yes, I understand all of this.WWGD said:The maximal width goes to 0 if we consider a net of Riemann sums. In order to compute the Riemann integral, we consider what happens as the width of partitions go to zero.
But any sum has a minimal width.
This is not what the Riemann integral is. One does not add up an infinite number of rectangles of zero width.Isaac0427 said:I know this discussion came off of my mentioning a Riemann sum. I meant a Riemann integral. My point was that in a Riemann integral, you add up an infinite amount of rectangles with a width of dx. dx, for the integral to have a nonzero value, couldn't be zero, but it is, by definition, less than .00000000001, .00000000000000000000000001, or any number of zeros you put before the one. You would need an infinite amount of zeros. So, I know it may not be the proper notation, but the width of the rectangles would essentially be .000...1. If the fact that it gets infinitely close to zero makes it equal to zero, then a Riemann integral would have a value of zero, which is not the case. If we were to take some fundamental length (in a coordinate system that has one), and have two particles of that size touch with a distance between them of zero, would they be in the same position? I say a fundamental size because the particle's centers would have no distance between them.
dx is also infinitesimally small, but if it equaled zero, then any integral or derivative would equal zero or be undefined.WWGD said:1-0.99999... _is_ indefinitely small, so must equal zero.
dx is not a number. It is a notation.Isaac0427 said:dx is also infinitesimally small, but if it equaled zero, then any integral or derivative would equal zero or be undefined.
I thought it represented an infinitely small change of x. I could be wrong, however.jbriggs444 said:dx is not a number. It is a notation.
Remember I said in previous posts that the length of the rectangles in a fixed Riemann sum is bounded below. You then consider what happens as this approaches 0 in the limit, but this, dealing with limits, is a separate issue.Isaac0427 said:dx is also infinitesimally small, but if it equaled zero, then any integral or derivative would equal zero or be undefined.
In the non-standard analysis, dx can be given meaning as an infinitesimal. In standard analysis, the dx in an integral is pure notation.Isaac0427 said:I thought it represented an infinitely small change of x. I could be wrong, however.
Isaac0427 said:Ok, I just talked to my math teacher. She explained it to me in person which helped. Thanks guys it really does help. I mostly understand it now, that as something gets so close to 1, for all practical purposes, it is equal to 1.
I know what a limit is. She explained how for practical purposes the limit .999... approaches 1 is considered .999... being equal to one.lavinia said:Ask your math teacher to explain to you what a limit is.
It is not "for practical purposes". It is exact. The limit of the sequence .9, .99, .999, ... does not approach 1. Successive terms in the sequence approach 1. The limit is 1. The notation .999... denotes the limit. Hence .999... is 1.Isaac0427 said:I know what a limit is. She explained how for practical purposes the limit .999... approaches 1 is considered .999... being equal to one.
The part where I bolded indicates you don't know what a limit is. Limits don't move and they don't "approach".Isaac0427 said:I know what a limit is. She explained how for practical purposes the limit .999... approaches 1 is considered .999... being equal to one.
HallsofIvy said:...
Restore that by adding ar^{n+1} to both sides:
S_n- a+ ar^n= r(a+ ar+ \cdot\cdot\cdot+ ar^{n-1})+ ar^{n+ 1}
...
You can prove that there is no difference at all because the 9's go forever. This proof may be your first encounter with a mathematical "proof by contradiction". Suppose you assume that there is any difference between .999999... and 1. Say it is over 0.00001 (a formal mathematical proof would use an arbitrarily small ε > 0). Now use enough 9's (0.999999999999) to show that there is less difference than that and that the difference will only decrease as you add more 9's. That contradicts to your original assumption that the difference is greater than 0.00001 (or ε>0). It doesn't matter how small your assumed difference is; you can add enough 9's to get closer to 1 and contradict that assumption. So it proves that there can be no difference between 0.99999... and 1.Isaac0427 said:Ok, I just talked to my math teacher. She explained it to me in person which helped. Thanks guys it really does help. I mostly understand it now, that as something gets so close to 1, for all practical purposes, it is equal to 1.
Yes, that was a typo. It should have been ar^{n+1} on both sides.gmax137 said:Going back to post #2
Maybe (probably) I'm being dense, but didn't you just add ar^n to the left and ar^{n+ 1} to the right?
Instead of thinking of .999... as a number, think of it as the limit of the sequence (.9,.99,.999,.9999,...) as n approaches infinity. That is what the 3 dots mean. The limit of that sequence as n approaches infinity is 1.Kegan said:i'm 11 i understand that 1 and .999 are two completely different numbers and in math you don't round numbers to get precise answers it just doesn't work like that
Isaac0427 said:Instead of thinking of .999... as a number
To be fair, Isaac didn't say it was not a number. He just suggested that it might be easier to see that 0.9999... is the limit of the sequence 0.9, 0.99, 0.999, ..., and that the limit of that sequence is also equal to 1, than it is to make the jump from the recurring decimal to 1 directly.micromass said:No. 0.99999... is a number.
MrAnchovy said:I don't think days of the week, which are analogous to integers mod 7 is a helpful analogy to the reals. And 2/2 is an expression which is equal to 1, it is not a valid representation of a rational number: the numerator and denominator must have no common factor.
Yes, maybe it is not the best example, but the point I wanted to make is that two things don't need to be strictly equal in order to be considered the same. It is more a " to the effects of what we are doing, these two expressions are equal" EDIT: Maybe non-trivial, i.e., non-identity isomorphisms would be a better example.MrAnchovy said:I don't think days of the week, which are analogous to integers mod 7 is a helpful analogy to the reals. And 2/2 is an expression which is equal to 1, it is not a valid representation of a rational number: the numerator and denominator must have no common factor.
I was going to write that one could consider the distinction between numerals or formulas on the one hand and numbers on the other. It seems that the distinction you want to make is between exemplars of an equivalence class and the class itself.WWGD said:Yes, maybe it is not the best example, but the point I wanted to make is that two things don't need to be strictly equal in order to be considered the same. It is more a " to the effects of what we are doing, these two expressions are equal" EDIT: Maybe non-trivial, i.e., non-identity isomorphisms would be a better example.
That was the point. I understand it is a number, but the value of the number can be thought of the limit of that sequence. IMO, this is easier to comprehend, but as you said, it's a mater of opinion.MrAnchovy said:To be fair, Isaac didn't say it was not a number. He just suggested that it might be easier to see that 0.9999... is the limit of the sequence 0.9, 0.99, 0.999, ..., and that the limit of that sequence is also equal to 1, than it is to make the jump from the recurring decimal to 1 directly.
yeah that does make more sense so what you saying is that .999... to infinity approaches the number 1 so close it is consideribly the same number is that what you are saying ?Isaac0427 said:Instead of thinking of .999... as a number, think of it as the limit of the sequence (.9,.99,.999,.9999,...) as n approaches infinity. That is what the 3 dots mean. The limit of that sequence as n approaches infinity is 1.
Here's another example: (1/0!)+(1/1!)+(1/2!)...=e, however if you stop the sequence at any value of n, the answer will not be e. Does this make sense?
yeah I Know i just forgot the...WWGD said:@Kegan It is not .999 , it is .9999... with an infinite string of 9s.
No, .999... doesn't approach 1 -- it is exactly equal to 1.Kegan said:yeah that does make more sense so what you saying is that .999... to infinity approaches the number 1 so close it is consideribly the same number is that what you are saying ?
Ok that makes sense the more 9's you have the closer it is to the number 1 ,and I'm only eleven and language arts is not my best subjectMark44 said:No, .999... doesn't approach 1 -- it is exactly equal to 1.
The limit of the sequence {.9, .99, .999, ...} is 1, which means the farther you go in the sequence, the closer a number in the sequence is to 1.
BTW, it's redundant to write "to infinity" after .999... The dots (an ellipsis) already means that the 9 digits repeat endlessly.
No, .999... is NOT a sequence. It is a number. The ... means that the value of the number is equal to the limit of the sequence (.9, .99, .999, ...). If it helps, you can think of it as the term of the sequence in which n=infinity. The value of this term, by definition, will be the limit of the sequence as n approaches infinity. If you do not understand this, look up limits.Kegan said:yeah that does make more sense so what you saying is that .999... to infinity approaches the number 1 so close it is consideribly the same number is that what you are saying ?
